Question

Find the points of local minima for the function $f\left(x\right)=-\frac{3}{4}{x}^4-8x^3-\frac{45}{2}{x}^2+105$

• A. $0$
• B. $-3$
• C. $-5$
• D. both $(b)$ and $(c)$

Answer 1

Answer: (B)

We have,
$f\left(x\right)=-\frac{3}{4}{x}^4-8x^3-\frac{45}{2}{x}^2+105$
$\Rightarrow f^{\prime }\left(x\right)=-3x^3-24x^2-45x$
$=-3x\left(x^2+8x+15\right)$
For local maximum or local minimum, we must have $f^{\prime }\left(x\right)=0$
$\Rightarrow -3x\left(x^2+8x+15\right)=0$
$\Rightarrow 3x\left(x+3\right)\left(x+5\right)=0$
$\Rightarrow x=0$, $-3$, $-5$
Thus, $x=0$, $x=-3$ and $x=-5$ are the possible points of local maxima or minima.
Now we test the function at each of these points.
We have, $f^{\prime \prime }(x)=-9x^2-48x-45$
At $x=0$ : We have,
$f^{\prime \prime }(0)=-45<0$
So, $x=0$ is point of local maximum.
At $x=-3$ : We have,
$f^{\prime \prime }(-3)=-9(-3{)}^2-48(-3)-45=18>0$
So, $x=-3$ is a point of local minimum.
At $x=-5$ : We have,
$f^{\prime \prime }(-5)=-9(-5{)}^2-48(-5)-45=-30<0$
So, $x=-5$ is point of local maximum.